6(3z-1)-4(z+4)=7(z+1)

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Solution for 6(3z-1)-4(z+4)=7(z+1) equation: 



6(3z-1)-4(z+4)=7(z+1)

We move all terms to the left:

6(3z-1)-4(z+4)-(7(z+1))=0

We multiply parentheses

18z-4z-(7(z+1))-6-16=0



We calculate terms in parentheses: -(7(z+1)), so:

7(z+1)

We multiply parentheses

7z+7

Back to the equation:

-(7z+7)



We add all the numbers together, and all the variables

14z-(7z+7)-22=0

We get rid of parentheses

14z-7z-7-22=0

We add all the numbers together, and all the variables

7z-29=0

We move all terms containing z to the left, all other terms to the right

7z=29

z=29/7

z=4+1/7

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